The wavelength of light with an energy of
is approximately
.
To find the wavelength
of light given its energy
, you can use the following formula:
![\[ E = h \cdot \\u \]](https://img.qammunity.org/2024/formulas/chemistry/high-school/uc05ymwnumftazyxmcjvcvi7dehgjzrjky.png)
Since
can also be expressed as
(where
is Planck's constant and
is the speed of light), we can set up the equation:
![\[ hc/\lambda = E \]](https://img.qammunity.org/2024/formulas/chemistry/college/bo4qg6xy3c8pof9krggvzpgt0ms46d4reo.png)
Solving for
:
![\[ \lambda = (hc)/(E) \]](https://img.qammunity.org/2024/formulas/chemistry/college/6ogrh4la83c892ep3fejkeslvzx25aksqt.png)
Given
,
, and
, you can substitute these values into the equation:
![\[ \lambda = \frac{(6.626 * 10^(-34) \, \text{J} \cdot \text{s}) \cdot (3.00 * 10^8 \, \text{m/s})}{1.50 * 10^(-18) \, \text{J}} \]](https://img.qammunity.org/2024/formulas/chemistry/college/7uv40714n8yp2bs0etw3atoazpglz8c89y.png)
Calculate this expression to find the wavelength
. The unit of wavelength will be meters.
![\[ \lambda \approx (1.987 * 10^(-25))/(1.50 * 10^(-18)) \, \text{m} \]](https://img.qammunity.org/2024/formulas/chemistry/college/w0d30x1hvtj6xkkfxwzgr87qwa3v8lcan5.png)
![\[ \lambda \approx 1.32 * 10^(-7) \, \text{m} \]](https://img.qammunity.org/2024/formulas/chemistry/college/ktq4vrgjclj5pej8mzpcnzcosb4mb0hfyn.png)
So, the wavelength of light with an energy of
is approximately
.
The probable question may be:
"What is the wavelength of light that has an energy of 1.50x
J "